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4- 1 4- 2 4- 3 4- 4 4- 5 Chemical Reactions and Chemical Equations Chemical Equations and Stoichiometry Chemical Reactions in Solution Determining the Limiting Reactant Other Practical Matters in Reaction Stoichiometry Chapter 4: Chemical Reactions
CHM 25 Spring 2007 Dr. Berk & Dr. Koel
Slide 1 of 46 CHM 25 SP07 LU Slide 2 of 46 CHM 25 SP07 LU Chemical Reactions and Chemical Equations
As reactants are converted to products we observe: Color change Precipitate formation Gas evolution Heat absorption or evolution Chemical Equations
Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.
Slide 4 of 46 CHM 25 SP07 LU Chemical evidence may be necessary.
Slide 3 of 46 CHM 25 SP07 LU Reaction of Phosphorus with Cl2 Chemical Equations
4 Al(s) + 3 O2(g) → 2 Al2O3(s) This equation means: 4 Al atoms + 3 O2 molecules ---give---> 2“ molecules” of Al2O3 Notice the stoichiometric coefficients and the physical states of the reactants and products.
Slide 5 of 46 CHM 25 SP07 LU 4 moles of Al + 3 moles of O2 ---give---> 2 moles of Al2O3
Slide 6 of 46 CHM 25 SP07 LU 2 CHM 25 SP07 LU Chemical Equations
Because of the principle of the conservation of matter, an equation Balancing Equations • Balancing chemical equations ensures that the same number of atoms of each element appear on both sides of the equation. • Many can be balanced by trial & error, some require special consideration. balanced. must be It must have the same number of atoms of the same kind on both sides.
Slide 7 of 46 Lavoisier, 1788 ___ Al(s) + ___ Br2(l) → ___ Al2Br6(s)
Slide 8 of 46 CHM 25 SP07 LU CHM 25 SP07 LU Balancing Equations An equation can be balanced only by adjusting the coefficients of formulas. Never introduce extraneous atoms to balance. Never change a formula for the purpose of balancing an equation.
Slide 9 of 46 CHM 25 SP07 LU Balancing Equation Strategy Balance elements that occur in only one compound on each side first. Balance free elements last. Balance unchanged polyatomics (or other groups of atoms) as groups. Fractional coefficients are acceptable and can be cleared at the end by multiplication.
Slide 10 of 46 CHM 25 SP07 LU Types of Chemical Reactions Reaction of a metal or non-metal with oxygen to give a metal oxide (MxOy). ___ Fe (s) + ___ O2 (g) → ___ Fe2O3 (s) Combustion Reactions. __C3H8(g) + __ O2(g) → __CO2(g) + __ H2O(g)
Slide 11 of 46 CHM 25 SP07 LU Slide 12 of 46 Types of Chemical Reactions Metathesis (Double Displacement or Exchange) Reactions. Precipitation Rxns. __Cu(NO3)2 (aq) + __Na3PO4 (aq) → __NaNO3 (aq) + __Cu3(PO4)2 (s) CHM 25 SP07 LU 3 CHM 25 SP07 LU Types of Chemical Reactions Acid-Base Reactions 4-2 Chemical Equations and Stoichiometry Stoichiometry includes all the quantitative relationships involving: atomic and formula masses chemical formulas. • Mole ratio is a central conversion factor. Oxidation Reduction Reactions
Slide 13 of 46 CHM 25 SP07 LU Slide 14 of 46 CHM 25 SP07 LU PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products? 454 g of NH4NO3 --> N2O + 2 H2O
STEP 2 Convert mass reactant moles reactant STEP 1 Write the balanced chemical equation
STEP 3 Convert moles reactant moles product __NH4NO3 → __N2O + __H2O Slide 15 of 46 CHM 25 SP07 LU Slide 16 of 46 CHM 25 SP07 LU 454 g of NH4NO3 --> N2O + 2 H2O
STEP 4 Convert moles product mass product GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
Mass reactant Mass product Called the THEORETICAL YIELD ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!
Slide 17 of 46 CHM 25 SP07 LU Moles reactant Stoichiometric factor Moles product Slide 18 of 46 CHM 25 SP07 LU 4 CHM 25 SP07 LU Theoretical, Actual and Percent Yield STEP 5 Calculate the percent yield
If you isolated only 131 g of N2O, what is the percent yield? This compares the theoretical (250. g) and 4-3 Chemical Reactions in Solution Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. Close contact between atoms, ions and molecules necessary for a reaction to occur.
Slide 20 of 46 CHM 25 SP07 LU actual (131 g) yields.
% yield =
Slide 19 of 46 actual yield theoretical yield • 100%
CHM 25 SP07 LU Terminology
In solution we need to define the SOLVENT the component whose physical state is preserved when solution forms SOLUTE the other solution component
Slide 21 of 46 CHM 25 SP07 LU An Ionic Compound, CuCl2, in Water Slide 22 of 46 CHM 25 SP07 LU Concentration of Solute Preparation of a Solution The amount of solute in a solution is given by its concentration.
Amount of solute (mol solute) Volume of solution (L)
Weigh the solid sample. Dissolve it in a volumetric flask partially filled with solvent. Carefully fill to the mark.
CHM 25 SP07 LU Slide 24 of 46 CHM 25 SP07 LU Molarity (M) = Concentration (M) = [ … ]
Slide 23 of 46 5 CHM 25 SP07 LU PROBLEM: Dissolve 5.00 g of NiCl22•6 H22O in PROBLEM: Dissolve 5.00 g of NiCl •6 H O in enough water to make 250 mL of solution. enough water to make 250 mL of solution. Calculate molarity.. Calculate molarity The Nature of a CuCl2 Solution: Ion Concentrations
CuCl2(aq) → Cu2+(aq) + 2 Cl-(aq) Step 1: Calculate moles of NiCl2• 2O 6H Step 2: Calculate molarity. If [CuCl2] = 0.30 M, then [Cu2+] = 0.30 M [Cl-] = 2 x 0.30 M
Slide 25 of 46 CHM 25 SP07 LU Slide 26 of 46 CHM 25 SP07 LU USING MOLARITY USING MOLARITY
What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/V this means that moles = M• V
Step 1: Calculate moles of acid required. moles = M• V
Slide 27 of 46 CHM 25 SP07 LU Step 2: Calculate mass of acid required.
Slide 28 of 46 CHM 25 SP07 LU Preparing a Solution by Dilution PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? H2 O Add water to the 3.0 M solution to lower its concentration to 0.50 M
3.0 M NaOH Concentrated 0.50 M NaOH Dilute Dilute the solution!
CHM 25 SP07 LU Slide 29 of 46 CHM 25 SP07 LU Slide 30 of 46 6 CHM 25 SP07 LU PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? How much water is added? The important point is that moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution Conclusion:
H2 O add _____ of water to 50.0 mL
3.0 M NaOH 0.50 M NaOH Dilute Cinitial •Vinitial = Cfinal •Vfinal Cinitial •Vinitial = Cfinal •Vfinal
Slide 31 of 46 CHM 25 SP07 LU Concentrated of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.
CHM 25 SP07 LU Slide 32 of 46 Solution Dilution
Mi Vi Mf Vf Reactions Involving a LIMITING REACTANT M= n V In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.
CHM 25 SP07 LU Slide 34 of 46 CHM 25 SP07 LU Mi × Vi = ni = nf = Mf × Vf Mf =
Slide 33 of 46 Mi × Vi Vf = Mi Vi Vf LIMITING REACTANTS PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? 2 Al (s) + 3 Cl2 (g) → Al2Cl6 (s)
Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.
mol Cl 2 3 = mol Al 2
Slide 36 of 46 CHM 25 SP07 LU Reactants Products 2 NO (g) + O2 (g) 2 NO2(g) Limiting reactant = ___________ Excess reactant = ____________
Slide 35 of 46 CHM 25 SP07 LU 7 CHM 25 SP07 LU Deciding on the Limiting Reactant Deciding on the Limiting Reactant 2 Al (s) + 3 Cl2 (g) → Al2Cl6 (s)
If 2 Al (s) + 3 Cl2 (g) → Al2Cl6 (s)
If mol Cl2 3 > mol Al 2 mol Cl2 3 < mol Al 2 There is not enough Al to use up all the Cl2 Limiting reagent = Al
Slide 37 of 46 CHM 25 SP07 LU There is not enough Cl2 to use up all the Al Limiting reagent = Cl2
Slide 38 of 46 CHM 25 SP07 LU Deciding on the Limiting Reactant
Step 2 of LR problem: Calculate moles of each reactant
We have 5.40 g of Al and 8.10 g of Cl2 Find mole ratio of reactants 2 Al (s) + 3 Cl2 (g) → Al2Cl6 (s)
mol Cl2 0.114 mol = mol Al 0.200 mol = 0.57 This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio.
Slide 39 of 46 CHM 25 SP07 LU Slide 40 of 46 CHM 25 SP07 LU Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? Limiting reactant = Cl22 Limiting reactant = Cl Base all calcs.. on Cl22 Base all calcs on Cl
1 mol Al2 Cl6 3 mol Cl 2 CALCULATIONS: calculate mass of Al2Cl6 expected. 2 Al (s) + 3 Cl2 (g) → Al2Cl6 (s)
Step 1: Calculate moles of Al2Cl6 expected based on LR. mass Al2Cl6 Step 2: Calculate mass of Al2Cl6 expected based on LR.
CHM 25 SP07 LU Slide 42 of 46 CHM 25 SP07 LU moles Cl2
Slide 41 of 46 8 CHM 25 SP07 LU How much of which reactant will remain when reaction is complete? Calculating Excess Al
2 Al + 3 Cl2 0.200 mol 0.200 mol products Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required. Then find how much Al is in excess.
Slide 43 of 46 CHM 25 SP07 LU 0.114 mol = LR 0.114 mol = LR
2 mol Al = 0.0760 mol Al req' d 3 mol Cl2 0.114 mol Cl2 • Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess
Slide 44 of 46 CHM 25 SP07 LU Consecutive Reactions, Simultaneous Reactions and Overall Reactions Multistep synthesis is often unavoidable. Reactions carried out in sequence are called consecutive reactions. When substances react independently and at the same time the reaction is a simultaneous reaction.
Slide 45 of 46 CHM 25 SP07 LU Overall Reactions and Intermediates The Overall Reaction is a chemical equation that expresses all the reactions occurring in a single overall equation. An intermediate is a substance produced in one step and consumed in another during a multistep synthesis. Slide 46 of 46 CHM 25 SP07 LU ...
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This note was uploaded on 10/24/2009 for the course CHEM 025 taught by Professor X during the Spring '06 term at Lehigh University .
- Spring '06